/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 72 A thin-walled, hollow spherical ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A thin-walled, hollow spherical shell of mass \(m\) and radius \(r\) starts from rest and rolls without slipping down a track (Fig. \(\mathbf{P 1 0 . 7 2}\) ). Points \(A\) and \(B\) are on a circular part of the track having radius \(R\). The diameter of the shell is very small compared to \(h_{0}\) and \(R,\) and the work done by rolling friction is negligible. (a) What is the minimum height \(h_{0}\) for which this shell will make a complete loop-the-loop on the circular part of the track? (b) How hard does the track push on the shell at point \(B,\) which is at the same level as the center of the circle? (c) Suppose that the track had no friction and the shell was released from the same height \(h_{0}\) you found in part (a). Would it make a complete loop-theloop? How do you know? (d) In part (c), how hard does the track push on the shell at point \(A,\) the top of the circle? How hard did it push on the shell in part (a)?

Short Answer

Expert verified
The minimum height \(h_0\) for the shell to make a complete loop is \(2.5R\). The track will push on the shell at point B with a force \(N\) equal to \(3mg\). Without friction, the shell won't be able to make a complete loop because it won't carry out a circular motion. The force with which the track pushes on the shell at point A (top of the circle) in case of slipping is zero, as there is no normal force at that point.

Step by step solution

01

Conservation of Energy

At the beginning, all the energy is potential energy. At the top of the loop, some of this energy has been converted into the kinetic energy, and the remaining is the potential energy. We can write the conservation of energy equation as: \(mgh_0 = \frac{1}{2}mv^2 + mgh\). Since the object is moving in a circular loop, the speed at the top must be at least \(v = \sqrt{gR}\) to keep contact with the track. Hence, \(mgh_0 = \frac{1}{2}m(2gR) + mg(2R)\). Solve this to find \(h_0\)
02

Calculating Normal Force at B

This can be found by applying Newton's second law in vertical direction at point B. The only forces acting on the shell are the weight \(mg\) and the normal force \(N\) from the track. These forces must provide the necessary centripetal force for the circular motion, hence \(N - mg = \frac{mv^2}{R}\). Since the shell started from rest and descended a height of \(h_0\), its speed is \(v = \sqrt{2g(h_0 - R)}\). Substitute this into the Newton's equation to find \(N\) at point B.
03

Loop-the-Loop without Friction

Without friction, the object cannot have a rolling motion, it would rather be in a slipping state. For an object to make a complete loop, it requires a minimum speed at the top so that it doesn't lose contact with the track. But without friction, there is no way to force the object into circular motion and it won't be able to make a full loop.
04

Normal Force at A without Rolling Motion

As the shell isn't rolling, the gravitational force provides the necessary centripetal force at the top, hence \(N = mg - mg = 0\). It means that there is no normal force acting on the shell at the top. The normal force differs from part (a) because in part (a), the shell is in a state of rolling and maintaining contact with the track so it does experience a normal force.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rolling Motion
Rolling motion is a combination of rotational and translational motion which occurs when an object like the hollow spherical shell in our exercise moves forward while also spinning around its axis. It is crucial to understand this combination, as it significantly affects how objects move along surfaces.

When a spherical shell rolls down a track, it is not just sliding. It maintains contact with the surface, and its entire motion can be broken down into:
  • Translational motion, where the center of the shell moves linearly down the track.
  • Rotational motion, where the shell spins around its own axis as it moves.
For rolling without slipping, the condition is that the velocity of the shell at the surface must match the velocity of the shell's center — effectively binding translational and rotational components. In the case of our shell, conservation of energy involves both kinetic energy from translational motion and kinetic energy from rotational motion along with potential energy. This leads to equations that must be satisfied to determine quantities like initial height for a complete loop.
Centripetal Force
Centripetal force is the force required to keep an object moving in a circular path and is directed towards the center around which the object is moving. This concept is key to understanding situations where objects like our spherical shell are moving through loop-the-loops.
For our shell, centripetal force must be provided as it moves through the curved section of the track. This force results from:
  • The gravitational pull of the shell's weight, which constantly acts downwards.
  • The normal force exerted by the track, which adjusts as needed to keep the shell in motion within the loop.
At the top of the loop, the velocity of the shell is dictated by the need to maintain a minimum speed to avoid losing contact. This minimum speed ensures that the gravitational force provides just enough centripetal force to maintain circular motion, calculated as \(v = \sqrt{gR}\) for the spherical shell. If the speed falls below this, the path cannot be maintained, leading to failure to complete the loop.
Normal Force
Normal force is a perpendicular force exerted by a surface against an object in contact with it. This force is fundamental when considering motion on tracks, especially in loop-the-loop scenarios.
In our exercise, the normal force varies at different points along the track. It plays a critical role in maintaining the spherical shell's contact with the track:
  • At point B, on the side of the loop, normal force adds up to the centripetal force needed to keep the shell on the circular path, calculated when balanced against gravitational force.
  • At the top of the loop (point A), if the shell is rolling, gravity itself needs to provide enough centripetal force, and in such cases, normal force may reduce or become negligible (zero), as the shell's weight does all the work to keep circular motion.
Understanding normal force allows us to determine how hard the track pushes on the shell at various points, contributing to the full picture of energy and motion conservation in our exercise. As such, it impacts whether or not the shell can complete the loop safely and successfully.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A large turntable with radius \(6.00 \mathrm{~m}\) rotates about a fixed vertical axis, making one revolution in \(8.00 \mathrm{~s}\). The moment of inertia of the turntable about this axis is \(1200 \mathrm{~kg} \cdot \mathrm{m}^{2}\). You stand, barefooted, at the rim of the turntable and very slowly walk toward the center, along a radial line painted on the surface of the turntable. Your mass is \(70.0 \mathrm{~kg}\). since the radius of the turntable is large, it is a good approximation to treat yourself as a point mass. Assume that you can maintain your balance by adjusting the positions of your feet. You find that you can reach a point \(3.00 \mathrm{~m}\) from the center of the turntable before your feet begin to slip. What is the coefficient of static friction between the bottoms of your feet and the surface of the turntable?

A wheel with radius \(0.0600 \mathrm{~m}\) rotates about a horizontal frictionless axle at its center. The moment of inertia of the wheel about the axle is \(2.50 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The wheel is initially at rest. Then at \(t=0 \mathrm{a}\) force \(F=(5.00 \mathrm{~N} / \mathrm{s}) t\) is applied tangentially to the wheel and the wheel starts to rotate. What is the magnitude of the force at the instant when the wheel has turned through 8.00 revolutions?

A \(12.0 \mathrm{~kg}\) box resting on a horizontal, frictionless surface is attached to a \(5.00 \mathrm{~kg}\) weight by a thin, light wire that passes over a frictionless pulley (Fig. E10.16). The pulley has the shape of a uniform solid disk of mass \(2.00 \mathrm{~kg}\) and diameter \(0.500 \mathrm{~m} .\) After the system is released, find (a) the tension in the wire on both sides of the pulley, (b) the acceleration of the box, and (c) the horizontal and vertical components of the force that the axle exerts on the pulley.

A \(2.20 \mathrm{~kg}\) hoop \(1.20 \mathrm{~m}\) in diameter is rolling to the right without slipping on a horizontal floor at a steady \(2.60 \mathrm{rad} / \mathrm{s}\). (a) How fast is its center moving? (b) What is the total kinetic energy of the hoop? (c) Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop; (iii) a point on the right side of the hoop, midway between the top and the bottom. (d) Find the velocity vector for each of the points in part (c), but this time as viewed by someone moving along with the same velocity as the hoop.

You live on a planet far from ours. Based on extensive communication with a physicist on earth, you have determined that all laws of physics on your planet are the same as ours and you have adopted the same units of seconds and meters as on earth. But you suspect that the value of \(g\), the acceleration of an object in free fall near the surface of your planet, is different from what it is on earth. To test this, you take a solid uniform cylinder and let it roll down an incline. The vertical height \(h\) of the top of the incline above the lower end of the incline can be varied. You measure the speed \(v_{\mathrm{cm}}\) of the center of mass of the cylinder when it reaches the bottom for various values of \(h .\) You plot \(v_{\mathrm{cm}}^{2}\) (in \(\mathrm{m}^{2} / \mathrm{s}^{2}\) ) versus \(h\) (in \(\mathrm{m}\) ) and find that your data lie close to a straight line with a slope of \(6.42 \mathrm{~m} / \mathrm{s}^{2}\). What is the value of \(g\) on your planet?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.